# Calculating effect of error percentages in trigonometry

Suppose a surveyor as 50 feet from the base of a tree, and the angle to the top is calculated to be $51.8^{\circ}$. How accurately must the angle be measured if the height error is less than 6%? The answer must be given in $d\theta$ radians.

Let the adjacent side= 50, the opposing side be x, and the ground angle be $51.8^{\circ}$. Therefore x is calculated as $$\frac{x}{50}= \tan{(51.8)}$$ $$x= 50~ \tan{(51.8)}$$ $$x= 63.5386$$

Since there is a +/- 6% error allowance, $$x+0.06= 63.5386+3.812=~67.3506\text{ feet for the upper bound}$$ $$x-0.06= 63.5386-3.812=~59.7266\text{ feet for the lower bound}$$

Using the inverse tangent function to determine the new angles necessary,

$$tan^{-1}\frac{67.3506}{50}= 53.410^{\circ} \text{ which is an increase of } 0.923 \text{ degrees}$$ $$tan^{-1}\frac{59.7266}{50}= 50.066^{\circ} \text{ which is an decrese of } 0.874 \text{ degrees}$$

As I have two different bounds, which is the the one to be used for the solution?

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Writing $x+0.06$ is not correct. It should either be $1.06x$ or $x+0.06x$. Some would write $x+6\%$ but that is not really right either. – Ross Millikan Apr 6 '11 at 1:11

If you want to be sure you are within $6\%$ and want symmetric bounds, you have to use the smaller bound. The difference is basically due to the second derivative of the tangent function-if the second derivative were zero the allowable error would be symmetric. You could use asymmetric bounds and say the measured angle must be $50.066^{\circ}$ to $53.410^{\circ}$, or $51.8^{\circ}+.923-.874$. I think most people would just round down and say $51.8^{\circ}\pm 0.8^{\circ}$, as your angle is just given to one place.